An affinely connected spacetime with a metric compatible connection can, in principle, have a non-vanishing anti-symmetric part; where the definition of the connection is given by defining parallel transport by
$$dA^{\mu} = - dx^{\sigma}\Gamma_{{\sigma}{\rho}}^{\mu}A^{\rho}$$
Now, in the light of Equivalence Principle, one can assert that at a point $\mathcal{P}$, one can always find a local inertial coordinate system with the coordinates of the point $\mathcal{P}$ being {$\xi^{\mu}$} and in this system, the parallel transported version of a vector $\vec{A}$ (whose coordinates at the point $\mathcal{P}$ are $\{A^\rho\}$) from $\mathcal{P}$ to a nearby point situated at coordinates $\{$$\xi^\mu+d\xi^\mu$$\}$ will also be $\{A^\rho\}$. Therefore, the components of the parallel transported version of $\vec{A}$ in a generic coordinate system with coordinates $\{x'^\mu\}$ will be
$\dfrac{\partial x'^{\mu}}{\partial \xi^{\rho}}\bigg|_{ \xi+d\xi}A^\rho$ and the components of the original vector $\vec{A}$ will be $\dfrac{\partial x'^{\mu}}{\partial \xi^{\rho}}\bigg|_{ \xi}A^\rho$. It is important to keep the distinction between coordinate transformation matrices of two different points because they will not generically be equal. Therefore, the difference in the coordinates of the parallel transported and the original vector in the generic coordinate system will become
$$ \begin{aligned} dA'^{\mu} &= \bigg(\dfrac{\partial x'^{\mu}}{\partial \xi^{\rho}}\bigg|_{ \xi+d\xi} - \dfrac{\partial x'^{\mu}}{\partial \xi^{\rho}}\bigg|_{ \xi}\bigg) A^{\rho}\\ &= \dfrac{\partial^2x'^{\mu}}{\partial \xi^{\sigma}\partial\xi^{\rho}}d\xi^{\sigma}A^{\rho} = \dfrac{\partial^2x'^{\mu}}{\partial \xi^{\sigma}\partial\xi^{\rho}}\dfrac{\partial \xi^{\sigma}}{\partial x'^{\kappa}} dx'^{\kappa}A^{\rho} = \dfrac{\partial^2x'^{\mu}}{\partial \xi^{\sigma}\partial\xi^{\rho}}\dfrac{\partial \xi^{\sigma}}{\partial x'^{\kappa}} \dfrac{\partial \xi^{\rho}}{\partial x'^{\nu}}dx'^{\kappa}A'^{\nu} \end{aligned} $$
Thus, under the light of Equivalence Principle, one can conclude that
$$\Gamma_{{\kappa}{\nu}}^{{'}{\mu}} = - \dfrac{\partial^2x'^{\mu}}{\partial \xi^{\sigma}\partial\xi^{\rho}}\dfrac{\partial \xi^{\sigma}}{\partial x'^{\kappa}} \dfrac{\partial \xi^{\rho}}{\partial x'^{\nu}}$$
The expression obtained is manifestly symmetric in the lower two indices of the connection and thus, the anti-symmetric part of the connection is zero. Doesn't this argument suffice to conclude that although there is no apriori mathematical reason to believe that torsion ($T_{{\mu}{\nu}}^{\lambda}:=-2\Gamma_{[{\mu}{\nu}]}^{\lambda}$) must be zero, under the light of Equivalence Principle, it is proven that the torsion must always vanish? I know that in General Relativity, torsion is indeed zero but I have read at many places that torsion can be incorporated into a theory of gravity and it is just an assumption made in General Relativity that torsion vanishes. I find this inappropriate as the Equivalence Principle dictates the vanishing torsion and one doesn't need to assume that. Also, I wonder how a theory allowing non-vanishing torsion can possibly accommodate Equivalence Principle - without which I think the considered theory should be of no merit to a Physicist.
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